Class 9 NCERT Solutions Physics Chapter 10 - Gravitation
Do you feel fascinated or intimidated while studying about gravitation in Physics? Whatever the case, you will find our NCERT Solutions for CBSE Class 9 Physics Chapter 10 Gravitation useful during revision. Revise the difference between weight and mass. Practise the expert solutions to understand the application of the law of gravitation to calculate the weight of an object on the Moon, Earth or other planets.
At TopperLearning, CBSE Class 9 Physics NCERT textbook solutions are available 24/7 along with other learning materials. Use these study materials by Science experts to grasp the important concept of the universal law of gravitation.
Gravitation Exercise 134
Solution 1
The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The gravitational force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
Consider two objects of masses m_{1} and m_{2} and let the distance between their centres be r. The gravitational force of attraction (F) acting between them is given by the universal law of gravitation as:
where, G is the universal gravitation constant given by:
Solution 2
Let M be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:
Gravitation Exercise 136
Solution 1
Gravity of the Earth attracts every object towards its centre. When an object is released from a height such that it falls towards the surface of the Earth under the influence of gravitational force alone, the motion of the object is called free fall.
Solution 2
During free fall of an object towards the earth, the magnitude of velocity of the falling object goes on increasing. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s^{2}.
The value of acceleration due to gravity changes from place to place but for most of the purposes it is taken to be 9.8 m/s^{2}.
Gravitation Exercise 138
Solution 1
S. No. |
Mass |
Weight |
I |
Mass is the quantity of matter contained in a body. |
Weight is the force of gravity acting on a body. |
II |
Mass of an object is a constant quantity. |
Weight of an object is not a constant quantity. It is different at different places. |
III |
It only has magnitude. |
It has magnitude as well as direction. |
IV |
Its SI unit is kilogram (kg). |
Its SI unit is the same as the SI unit of force, i.e., Newton (N). |
V |
Mass of an object is never zero. |
Weight of an object is zero at the centre of Earth. |
Concept Insight Weight is a force but mass is not.
Solution 2
Let M_{E} be the mass of the Earth and m be the mass of an object on the surface of the Earth. Let R_{E} be the radius of the Earth. According to the universal law of gravitation, weight W_{E} of the object on the surface of the Earth is given by,
Let M_{M }and R_{M} be the mass and radius of the Moon. Then, according to the universal law of gravitation, weight W_{M} of the same object on the surface of the Moon is given by:
Therefore, weight of an object on the moon is th of its weight on the Earth.
Concept Insight Gravitational force of the moon is th that of the earth.
Gravitation Exercise 141
Solution 1
It is difficult to hold a school bag having a thin strap because the pressure exerted on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulders is very large.
Solution 2
Whenever an object is immersed in a fluid, either partially or fully, it experiences an upward force by the fluid. This force is called buoyant force or upthrust and the property due to which a fluid exerts this upthrust on an object placed in it, is called buoyancy.
Solution 3
If the density of an object is more than the density of water, then it sinks in water. This is because the buoyant force acting on the object is less than the force of gravity acting on it. On the other hand, if the density of an object is less than the density of water, then it floats on the surface of water. This is because the buoyant force acting on the object is greater than the force of gravity acting on it.
Concept Insight -Density of an object is a key factor to decide whether it'll float or sink in a liquid.
Gravitation Exercise 142
Solution 1
When you weigh your body, an upward force (due to air) acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.
Concept Insight -The upthrust of air acting on the body reduces the weight of the body.
Solution 2
We know that true weight = apparent weight + upthrust. The cotton bag is heavier than the iron bar. This is due to the reason, that the bag of cotton which has more volume (as it has less density) than the iron bar (which has more density), experiences more upthrust due to air.
Gravitation Exercise 143
Solution 1
According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance (r) between them, i.e.
Hence, if the distance is reduced to half, then the gravitational force would become four times.
Solution 2
It is true that gravitational force acts on all objects in proportion to their masses. But a heavy object does not fall faster than a light object. This is because of the reason that
As Force Mass, therefore, acceleration is constant for a body of any mass.
Solution 3
According to the universal law of gravitation, gravitational force between two objects of masses M and m at a distance r from each other is given by:
Let mass of the object be represented by m = 1 kg
Universal gravitational constant, G = 6.7 × 10^{-11} Nm^{2} kg^{-2}
Since the object is on the surface of the Earth, r = radius of the Earth (R)
r = R = 6.4 × 10^{6} m
Solution 4
The Earth attracts the Moon with a force which is same as the force with which the moon attracts the Earth because according to Newton's third law of motion, force of action and reaction are always equal and opposite. So, the force of attraction of Earth on Moon is equal and opposite to the force of attraction of Moon on Earth.
Solution 5
The Earth and the Moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the Moon. Hence, it accelerates at a rate much lesser than the acceleration rate of the Moon towards the Earth. For this reason, we do not see the Earth moving towards the Moon.
Concept Insight - For the same force, an object with greater mass obtains less acceleration as compared to an object with lesser mass.
Gravitation Exercise 144
Solution 6
(ii) F is inversely proportional to the square of the distance between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value.
Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.
(iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.
Solution 7
The universal law of gravitation is important because it accounts for the force that binds us to the Earth, motion of planets around the Sun, motion of the Moon and other artificial satellites around the Earth, tides due to the Moon and the Sun and many other phenomena.
Solution 8
When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. This force produces a uniform acceleration in the objects, which is called acceleration of free fall or acceleration due to gravity. Its value is 9.8 m/s^{2}.
Concept Insight - The value of acceleration of free fall is 9.8 m/s^{2}, which is constant for all objects (irrespective of their masses).Solution 9
Gravitational force between the Earth and an object is known as Earth's gravity or weight of the object.
Solution 10
Weight of a body on the earth is given by:
W = m g
where,
m = Mass of the body
g = Acceleration due to gravity
The value of g is greater at poles than at the equator. Therefore, the same mass of gold weighs lesser at the equator than at the poles. Hence, Amit's friend will not agree with the weight of the gold bought.
Solution 11
A sheet of paper has more surface area than a crumpled ball of paper. So, the resistance offered by air to a sheet of paper falling through it is more than the resistance offered to a falling crumpled ball of paper. This decreases the speed of the sheet of paper and hence it falls slower than the crumpled ball.
Solution 12
Weight = Mass × Acceleration
Acceleration due to gravity on earth, g_{e }= 9.8 m/s^{2}
Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N
Acceleration due to gravity on Moon,
Therefore, weight of the same object on the Moon =
Concept Insight Weight of an object on Moon = of the weight of the object on Earth
Solution 13
(i) For the upward motion of the ball, we use the equation:
v^{2} - u^{2} = 2 g h
where,
u = Initial velocity of the ball = 49 m/s (Given)
v = Final velocity of the ball = 0 (At the highest point)
h = Maximum height attained by the ball
g = Acceleration due to gravity = -9.8 m/s^{2} (Ball goes up)
Putting the values, we get
0 - (49)^{2} = 2 × (-9.8) × h
v = u + g t
Putting the values, we get
0 = 49 + (-9.8) × t
0 = 49 - 9.8 t
49 = 9.8 t
Concept Insight - When a body is thrown vertically upwards, its velocity decreases, so the acceleration due to gravity g is taken as negative.
Solution 14
Initial velocity of the stone, u = 0
Final velocity of the stone, v = ?
Height of the tower, h = 19.6 m
Acceleration due to gravity, g = 9.8 ms^{-2}
For a freely falling body:
v^{2 }- u^{2} = 2 g h
v^{2} - 0^{2} = 2 × 9.8 × 19.6
v^{2} = 19.6 × 19.6 = (19.6)^{2 }
v^{2} = 384.16
v = 19.6 ms^{-1}
Hence, the velocity of the stone just before touching the ground is 19.6 ms^{-1}.
Concept Insight When a body is falling vertically downwards, its velocity increases, so the acceleration due to gravity g is taken as positive.
Solution 15
Given:
Initial velocity of the stone, u = 40 m/s
Final velocity of the stone, v = 0 (At the highest point)
Maximum height reached by the stone, h = ?
Acceleration due to gravity, g = -10 ms^{-2} (Stone goes up)
Using the equation of motion:
v^{2} - u^{2} = 2 g h
0 - (40)^{2} = 2 × (-10) × h
0 - 1600 = - 20 h
-1600 = -20 h
1600 = 20 h
Thus, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey = 0 (since final position coincides with the initial position)
Concept Insight Distance is the length of the actual path covered by an object, while displacement is the shortest distance between the initial and final position of the object.
Solution 16
According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by:
where,
M_{Sun} = Mass of the Sun = 2 × 10^{30} kg
M_{Earth} = Mass of the Earth = 6 × 10^{24} kg
R = Average distance between the Earth and the Sun = 1.5 × 10^{11} m
G = Universal gravitational constant = 6.7 × 10^{-11} Nm^{2 }kg^{-2}
Substituting the values, we get
Solution 17
Let the two stones meet after a time t.
(i) For the stone dropped from the top of the tower:
Initial velocity, u = 0
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 ms^{-2}
From the equation of motion,
(ii) For the stone thrown upwards:
Initial velocity, u = 25 ms^{-1 }
Let the displacement of the stone from the ground in time t be s'.
Acceleration due to gravity, g = -9.8 ms^{-2}
From the equation of motion,
Concept Insight Choose the equation of motion wisely to minimize the number of steps in calculations.
Solution 18
The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height.
Let the initial velocity of the ball be u.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g = -9.8 ms^{-2}
From the equation of motion, v = u + gt, we get,
0 = u + (-9.8) × 3
0 = u - 29.4
u = 29.4 ms^{-1}
Hence, the ball was thrown upwards with a velocity of 29.4 ms ^{-1}.
(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 ms ^{-1}
Final velocity, v = 0
Acceleration due to gravity, g = -9.8 m s ^{-2}
Concept Insight Choose the equation of motion wisely to minimize the number of steps in calculations.
In this case,
Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by its position during the downward journey in 4 s - 3 s = 1 s.
Concept Insight - Choose the equation of motion wisely to minimize the number of steps in calculations.This means that the ball is 39.2 m (44.1 m - 4.9 m) above the ground after 4 seconds.
Gravitation Exercise 145
Solution 19
Solution 20
In case of a block of plastic, the upward buoyant force is greater than the weight of the object. The large buoyant force on the block is due to its density being smaller than that of water. Due to the larger buoyant force, the block of plastic comes up when released under water.
Concept Insight - A body floats when its weight is lesser than the upward buoyant force acting on it.
Solution 21
If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.
Solution 22
As the packet is fully submerged in water,
Mass of water displaced by the packet = volume of the packet x density of water
= 350 cm^{3} × 1 g/cm^{3} = 350 g
Concept Insight - Density of an object is a key factor in deciding whether the object will sink or float.